Types of Valuations

We define two important properties of valuations, both of which apply to equivalence classes of valuations (i.e., the property holds for $ \left\vert \cdot \right\vert{}$ if and only if it holds for a valuation equivalent to $ \left\vert \cdot \right\vert{}$).

Definition 15.2.1 (Discrete)   A valuation $ \left\vert \cdot \right\vert{}$ is if there is a $ \delta>0$ such that for any $ a\in K$

$\displaystyle 1-\delta < \left\vert a\right\vert < 1+\delta \implies \vert a\vert=1.

Thus the absolute values are bounded away from $ 1$.

To say that $ \left\vert \cdot \right\vert{}$ is discrete is the same as saying that the set

$\displaystyle G=\bigl\{
\log\left\vert a\right\vert : a \in K, a\neq 0
\bigr\} \subset \mathbf{R}

forms a discrete subgroup of the reals under addition (because the elements of the group $ G$ are bounded away from 0).

Proposition 15.2.2   A nonzero discrete subgroup $ G$ of $ \mathbf{R}$ is free on one generator.

Proof. Since $ G$ is discrete there is a positive $ m\in G$ such that for any positive $ x\in G$ we have $ m\leq x$. Suppose $ x\in G$ is an arbitrary positive element. By subtracting off integer multiples of $ m$, we find that there is a unique $ n$ such that

$\displaystyle 0\leq x-nm <m.

Since $ x-nm\in G$ and $ 0<x-nm<m$, it follows that $ x-nm=0$, so $ x$ is a multiple of $ m$. $ \qedsymbol$

By Proposition 15.2.2, the set of $ \log\left\vert a\right\vert$ for nonzero $ a\in K$ is free on one generator, so there is a $ c<1$ such that $ \left\vert a\right\vert$, for $ a\neq 0$, runs precisely through the set

$\displaystyle c^\mathbf{Z}= \{c^m : m\in \mathbf{Z}\}$

(Note: we can replace $ c$ by $ c^{-1}$ to see that we can assume that $ c<1$).

Definition 15.2.3 (Order)   If $ \left\vert a\right\vert = c^m$, we call $ m=\ord (a)$ the of $ a$.

Axiom (2) of valuations translates into

$\displaystyle \ord (ab) = \ord (a) + \ord (b).

Definition 15.2.4 (Non-archimedean)   A valuation $ \left\vert \cdot \right\vert{}$ is if we can take $ C=1$ in Axiom (3), i.e., if

$\displaystyle \vert a + b\vert \leq \max\bigl\{\vert a\vert,\vert b\vert\bigr\}.$ (15.3)

If $ \left\vert \cdot \right\vert{}$ is not non-archimedean then it is .

Note that if we can take $ C=1$ for $ \left\vert \cdot \right\vert{}$ then we can take $ C=1$ for any valuation equivalent to $ \left\vert \cdot \right\vert{}$. To see that (15.2.1) is equivalent to Axiom (3) with $ C=1$, suppose $ \vert b\vert\leq \vert a\vert$. Then $ \vert b/a\vert\leq 1$, so Axiom (3) asserts that $ \vert 1+b/a\vert\leq 1$, which implies that $ \vert a+b\vert \leq \vert a\vert = \max\{\vert a\vert,\vert b\vert\}$, and conversely.

We note at once the following consequence:

Lemma 15.2.5   Suppose $ \left\vert \cdot \right\vert{}$ is a non-archimedean valuation. If $ a, b\in K$ with $ \vert b\vert<\vert a\vert$, then $ \vert a+b\vert=\vert a\vert.

Proof. Note that $ \vert a+b\vert\leq \max\{\vert a\vert,\vert b\vert\} = \vert a\vert$, which is true even if $ \vert b\vert=\vert a\vert$. Also,

$\displaystyle \vert a\vert = \vert(a+b) - b\vert \leq \max\{\vert a+b\vert, \vert b\vert\} = \vert a+b\vert,

where for the last equality we have used that $ \vert b\vert<\vert a\vert$ (if $ \max\{\vert a+b\vert,\vert b\vert\} = \vert b\vert$, then $ \vert a\vert\leq \vert b\vert$, a contradiction).

$ \qedsymbol$

Definition 15.2.6 (Ring of Integers)   Suppose $ \left\vert \cdot \right\vert$ is a non-archimedean absolute value on a field $ K$. Then

$\displaystyle \O = \{a\in K : \vert a\vert\leq 1\}

is a ring called the of $ K$ with respect to $ \left\vert \cdot \right\vert{}$.

Lemma 15.2.7   Two non-archimedean valuations $ \left\vert \cdot \right\vert{}_1$ and $ \left\vert \cdot \right\vert{}_2$ are equivalent if and only if they give the same $ \O$.

We will prove this modulo the claim (to be proved later in Section 16.1) that valuations are equivalent if (and only if) they induce the same topology.

Proof. Suppose suppose $ \left\vert \cdot \right\vert{}_1$ is equivalent to $ \left\vert \cdot \right\vert{}_2$, so $ \left\vert \cdot \right\vert{}_1 = \left\vert \cdot \right\vert{}_2^c$, for some $ c>0$. Then $ \left\vert c\right\vert _1 \leq 1$ if and only if $ \left\vert c\right\vert _2^c \leq 1$, i.e., if $ \left\vert c\right\vert _2 \leq 1^{1/c}=1$. Thus $ \O _1 = \O _2$.

Conversely, suppose $ \O _1 = \O _2$. Then $ \vert a\vert _1<\vert b\vert _1$ if and only if $ a/b\in \O _1$ and $ b/a\not\in \O _1$, so

$\displaystyle \vert a\vert _1<\vert b\vert _1 \iff \vert a\vert _2 < \vert b\vert _2.$ (15.4)

The topology induced by $ \vert$  $ \vert _1$ has as basis of open neighborhoods the set of open balls

$\displaystyle B_1(z,r) = \{x \in K : \vert x-z\vert _1<r \},

for $ r>0$, and likewise for $ \vert$  $ \vert _2$. Since the absolute values $ \vert b\vert _1$ get arbitrarily close to 0, the set $ \mathcal{U}$ of open balls $ B_1(z,\vert b\vert _1)$ also forms a basis of the topology induced by $ \vert$  $ \vert _1$ (and similarly for $ \vert$  $ \vert _2$). By (15.2.2) we have

$\displaystyle B_1(z,\vert b\vert _1) = B_2(z,\vert b\vert _2),

so the two topologies both have $ \mathcal{U}$ as a basis, hence are equal. That equal topologies imply equivalence of the corresponding valuations will be proved in Section 16.1. $ \qedsymbol$

The set of $ a\in \O$ with $ \vert a\vert<1$ forms an ideal $ \mathfrak{p}$ in $ \O$. The ideal $ \mathfrak{p}$ is maximal, since if $ a\in \O$ and $ a\not\in\mathfrak{p}$ then $ \vert a\vert=1$, so $ \vert 1/a\vert = 1/\vert a\vert = 1$, hence $ 1/a\in \O$, so $ a$ is a unit.

Lemma 15.2.8   A non-archimedean valuation $ \left\vert \cdot \right\vert{}$ is discrete if and only if $ \mathfrak{p}$ is a principal ideal.

Proof. First suppose that $ \left\vert \cdot \right\vert{}$ is discrete. Choose $ \pi \in \mathfrak{p}$ with $ \vert\pi\vert$ maximal, which we can do since

$\displaystyle S=\{\log\vert a\vert : a \in \mathfrak{p}\} \subset (-\infty,1],

so the discrete set $ S$ is bounded above. Suppose $ a\in\mathfrak{p}$. Then

$\displaystyle \left\vert\frac{a}{\pi}\right\vert = \frac{\left\vert a\right\vert}{\left\vert\pi \right\vert} \leq 1,

so $ a/\pi\in \O$. Thus

$\displaystyle a = \pi \cdot \frac{a}{\pi} \in \pi \O .$

Conversely, suppose $ \mathfrak{p}=(\pi)$ is principal. For any $ a\in\mathfrak{p}$ we have $ a=\pi b$ with $ b\in\O$. Thus

$\displaystyle \vert a\vert = \vert\pi\vert\cdot \vert b\vert \leq \vert\pi\vert < 1.

Thus $ \{\vert a\vert : \vert a\vert<1\}$ is bounded away from $ 1$, which is exactly the definition of discrete. $ \qedsymbol$

Example 15.2.9   For any prime $ p$, define the $ p$-adic valuation $ \left\vert \cdot \right\vert{}_p:\mathbf{Q}\to\mathbf{R}$ as follows. Write a nonzero $ \alpha\in K$ as $ p^n\cdot \frac{a}{b}$, where $ \gcd(a,p)=\gcd(b,p)=1$. Then

$\displaystyle \left\vert p^n\cdot \frac{a}{b}\right\vert _p := p^{-n} = \left(\frac{1}{p}\right)^{n}.$

This valuation is both discrete and non-archimedean. The ring $ \O$ is the local ring

$\displaystyle \mathbf{Z}_{(p)} = \left\{\frac{a}{b}\in\mathbf{Q}: p\nmid b\right\},

which has maximal ideal generated by $ p$. Note that $ \ord (p^n\cdot \frac{a}{b}) = p^n.$

We will using the following lemma later (e.g., in the proof of Corollary 16.2.4 and Theorem 15.3.2).

Lemma 15.2.10   A valuation $ \left\vert \cdot \right\vert{}$ is non-archimedean if and only if $ \vert n\vert\leq
1$ for all $ n$ in the ring generated by $ 1$ in $ K$.

Note that we cannot identify the ring generated by $ 1$ with  $ \mathbf{Z}$ in general, because $ K$ might have characteristic $ p>0$.

Proof. If $ \left\vert \cdot \right\vert{}$ is non-archimedean, then $ \vert 1\vert\leq 1$, so by Axiom (3) with $ a=1$, we have $ \vert 1+1\vert\leq 1$. By induction it follows that $ \vert n\vert\leq

Conversely, suppose $ \vert n\vert\leq
1$ for all integer multiples $ n$ of $ 1$. This condition is also true if we replace $ \left\vert \cdot \right\vert{}$ by any equivalent valuation, so replace $ \left\vert \cdot \right\vert{}$ by one with $ C\leq 2$, so that the triangle inequality holds. Suppose $ a\in K$ with $ \vert a\vert\leq 1$. Then by the triangle inequality,

$\displaystyle \left\vert 1+a\right\vert^n$ $\displaystyle = \left\vert(1+a)^n\right\vert$    
$\displaystyle \leq$ $\displaystyle \sum_{j=0}^n \left\vert\binom{n}{j}\right\vert \left\vert a\right\vert$    
$\displaystyle \leq$ $\displaystyle 1 + 1 + \cdots + 1 = n.$    

Now take $ n$th roots of both sides to get

$\displaystyle \left\vert 1+a\right\vert \leq \sqrt[n]{n},$

and take the limit as $ n\to \infty$ to see that $ \left\vert 1+a\right\vert \leq 1$. This proves that one can take $ C=1$ in Axiom (3), hence that $ \left\vert \cdot \right\vert{}$ is non-archimedean. $ \qedsymbol$

William Stein 2004-05-06